a,
Giả sử $C$ dư
$C+2H_2\xrightarrow{{Ni, t^o}} CH_4$
$n_{H_2}=\dfrac{16,8}{22,4}=0,75(mol)$
Đặt $x$ là số mol $H_2$ phản ứng
$\to$ dư $0,75-x$ mol $H_2$
$n_{CH_4}=\dfrac{n_{H_2\text{pứ}}}{2}=0,5x(mol)$
$\overline{M}_{\text{hh spu}}=4,5.2=9$
$\to 2(0,75-x)+16.0,5x=9(0,75-x+0,5x)$
$\to x=0,5$
Vậy $H=\dfrac{0,5.100}{0,75}=66,67\%$
b,
Hỗn hợp spu gồm: $H_2$ ($0,25$), $CH_4$ ($0,25$ mol)
$CH_4+2O_2\xrightarrow{{t^o}} CO_2+2H_2O$
$2H_2+O_2\xrightarrow{{t^o}} 2H_2O$
$\to n_{CO_2}=n_{CH_4}=0,25(mol)$
$n_{NaOH}=\dfrac{200.1,1.8\%}{40}=0,44(mol)$
$\dfrac{n_{NaOH}}{n_{CO_2}}=1,76$
$\to$ tạo $NaHCO_3$ ($a$), $Na_2CO_3$ ($b$)
$NaOH+CO_2\to NaHCO_3$
$2NaOH+CO_2\to Na_2CO_3+H_2O$
$\to \begin{cases} a+b=0,25\\ a+2b=0,44\end{cases}$
$\to \begin{cases} a=0,06\\ b=0,19\end{cases}$
$m_{NaHCO_3}=0,06.84=5,04g$
$m_{Na_2CO_3}=0,19.106=20,14g$